Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, par[100005][20], dep[100005];
vector<int> v[100005], have[100005];
struct node {
  vector<int> id;
  void insert(int num) {
    id.push_back(num);
    sort(id.begin(), id.end());
    if ((int)id.size() > 10) id.pop_back();
  }
} val[100005][20];
node merge(node a, node b) {
  node ret;
  int idx = 0, idx2 = 0;
  while (idx < (int)a.id.size() && idx2 < (int)b.id.size() &&
         (int)ret.id.size() <= 10) {
    if (a.id[idx] < b.id[idx2])
      ret.id.push_back(a.id[idx]), ++idx;
    else
      ret.id.push_back(b.id[idx2]), ++idx2;
  }
  while (idx < (int)a.id.size() && (int)ret.id.size() <= 10)
    ret.id.push_back(a.id[idx]), ++idx;
  while (idx2 < (int)b.id.size() && (int)ret.id.size() <= 10)
    ret.id.push_back(b.id[idx2]), ++idx2;
  return ret;
}
node get(int p, int gap) {
  node ret;
  for (int i = 0; (1 << i) <= gap; i++) {
    if (gap & (1 << i)) {
      ret = merge(ret, val[p][i]);
      p = par[p][i];
    }
  }
  return ret;
}
int get_par(int p, int gap) {
  for (int i = 0; (1 << i) <= gap; i++) {
    if (gap & (1 << i)) {
      p = par[p][i];
    }
  }
  return p;
}
int LCA(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  a = get_par(a, dep[a] - dep[b]);
  if (a == b) return a;
  for (int i = 19; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      a = par[a][i], b = par[b][i];
    }
  }
  return par[a][0];
}
void dfs(int now, int pre) {
  par[now][0] = pre;
  for (int i = 0; i < have[now].size(); i++) {
    val[now][0].insert(have[now][i]);
  }
  for (int i = 0; i < v[now].size(); i++) {
    int nxt = v[now][i];
    if (nxt == pre) continue;
    dep[nxt] = dep[now] + 1;
    dfs(nxt, now);
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int a, b, i = 1; i <= n - 1; i++) {
    scanf("%d %d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int C, i = 1; i <= m; i++) {
    scanf("%d", &C);
    have[C].push_back(i);
  }
  dfs(1, 1);
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      par[i][j] = par[par[i][j - 1]][j - 1];
      val[i][j] = merge(val[i][j - 1], val[par[i][j - 1]][j - 1]);
    }
  }
  for (int a, b, c, ite = 1; ite <= q; ite++) {
    scanf("%d %d %d", &a, &b, &c);
    int an = LCA(a, b);
    node ret = get(a, dep[a] - dep[an]);
    node ret2 = get(b, dep[b] - dep[an] + 1);
    ret = merge(ret, ret2);
    if (ret.id.empty()) {
      puts("0");
      continue;
    }
    int sz = min((int)ret.id.size(), c);
    printf("%d", sz);
    for (int i = 0; i < sz; i++) {
      printf(" %d", ret.id[i]);
    }
    puts("");
  }
}